Question 1139071: A man walks a distance of 10km from his house to the church on a bearing of 56 degrees . He then walked to the market a distance of 5km on a bearing of 146 degrees .Calculate the bearing of the house from the market.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! my worksheet is shown below:
the bearing of 56 degrees is angle BAE.
bearings are the angles made from the north line going clockwise.
line AD and line EB are parallel and congruent.
line EA and BD are parallel and congruent.
this forms rectangle EBDA.
line AD is the diagonal to this rectangle.
it is the distance from point A to point B.
point A is the house.
point B is the church.
once at the church, the market is 5 km away at a bearing of 146 degrees.
that will be point C.
angle FBC is the angle that represents bearing of 146 degrees.
angle FBC is composed of angle FBG which is 90 degrees, and angle GBC which is 56 degrees.
since angle GBD is 90 degrees, angle CBD is equal to 34 degrees.
since AB is a diagonal of rectangle EBDA, then angle EAB is equal to angle DBA, each being 56 degrees.
what you have is:
the house at point A and the church at point B and the market at point C form triangle ABC.
this triangle has a 90 degree angle at point B which is angle ABC.
that makes triangle ABC a right triangle with leg AB equal to 10 kilometers and leg BC equal to 5 kilometers.
the angle you want is angle BAC which is labeled as angle A1.
since triangle ABC is a right triangle, then tan(A1) = opp/adj = 5/10.
arctan(5/10) is equal to 26.56505118 degrees.
the bearing from point A to point C is therefore 56 + 26.56505118 degrees = 82.56505118 degrees.
that should be your solution.
the bearing is the addition of angle EAB and angle BAC.
if angle ABC was, for some reason, not equal to 90 degrees, then you could still have found angle BAC through use of the law of cosines.
any questions, email dtheophilis@gmail.com.
note that lines AE and FD are vertical.
also note that lines EG and HD are horizontal.
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